Question

A conducting rod AC of length $ 4\,l $ is rotated about a point O in a uniform magnetic field B directed into the paper $ AO=l $ and $ OC=3\,l, $ then

• A. $ {{V}_{A}}-{{V}_{O}}=\frac{B\omega \,{{l}^{2}}}{2} $
• B. $ {{V}_{O}}-{{V}_{C}}=\frac{9}{2}B\omega \,{{l}^{2}} $
• C. $ {{V}_{A}}-{{V}_{C}}=4B\omega \,{{l}^{2}} $
• D. $ {{V}_{C}}-{{V}_{O}}=\frac{9}{2}B\omega \,{{l}^{2}} $

Answer 1

Answer: (C)

$ {{V}_{0}}-{{V}_{A}}=\frac{B\,\omega \,I}{2}\times I $ $ =\frac{B\,\omega \,{{I}^{2}}}{2} $ $ {{V}_{0}}-{{V}_{A}}=\frac{B\,\omega 3I}{2}\times 3I $ $ =\frac{9B\,\omega \,{{I}^{2}}}{2} $ $ {{V}_{A}}-{{V}_{c}}=4B\omega \,{{I}^{2}} $